TransferCredit.org Guide to AP® Calculus BC

Integration adds up a rate of change to find the total change, the reverse of differentiation. This unit builds the definite integral from Riemann sums, links it to antiderivatives through the Fundamental Theorem of Calculus, and adds the BC techniques: integration by parts, linear partial fractions and improper integrals. It is one of the two heaviest units on the BC exam.

Key terms, with examples

accumulation of change and its units

The area between a rate graph and the horizontal axis is the accumulated change, positive where the rate is positive. Units: (rate unit) × (input unit).

Example: A heater runs at a constant 3 kilowatts for 5 hours: the rectangle's area is 3 × 5 = 15 kilowatt-hours of energy.

definite integral as a limit of Riemann sums

Riemann sums Σ f(xᵢ*)Δxᵢ add value-times-width products, xᵢ* in piece i. As the widest Δxᵢ shrinks to 0, they approach ∫ from a to b of f(x) dx.

Example: Right sums for 4 − x² on [0, 2]: 3 with n = 2, 4.25 with n = 4, ≈ 5.29 with n = 100, nearing 16/3.

accumulation function

g(x) = ∫ from a to x of f(t) dt, with a fixed and x the input: g(x) is the signed area under f from a to x.

Example: g(x) = ∫ from 2 to x of 4 dt = 4x − 8: g(5) = 12, a 3-by-4 rectangle, and g(2) = 0.

indefinite integral and constant of integration

Every antiderivative of f has the form F(x) + C, written ∫ f(x) dx, where F′ = f and C is a constant. Check by differentiating.

Example: ∫ 4x³ dx = x⁴ + C: x⁴ + 7 and x⁴ − 1 both have derivative 4x³.

u-substitution

Undo the chain rule: pick u, an inner expression whose derivative du = u′(x) dx sits in the integrand (up to a constant). Integrate in u, then put x back.

Example: ∫ cos(5x) dx: u = 5x and du = 5 dx, so the integral is (1/5)sin(5x) + C.

integration by parts

For a product, pick u to differentiate and dv to integrate: ∫ u dv = uv − ∫ v du. It reverses the product rule; choose u so that ∫ v du is simpler.

Example: ∫ x eˣ dx: u = x and dv = eˣ dx give xeˣ − ∫ eˣ dx = xeˣ − eˣ + C.

improper integral with an infinite limit

Replace ∞ with b and take a limit: ∫ from a to ∞ of f(x) dx = lim b→∞ ∫ from a to b of f(x) dx. A finite limit means it converges.

Example: ∫ from 0 to ∞ of 1/(x + 2)² dx = lim b→∞ (1/2 − 1/(b + 2)) = 1/2, so it converges.

left and right Riemann sums

Add rectangle areas, width times height, using each subinterval's left endpoint or right endpoint. Equal widths: L = Δx[f(x₀) + … + f(xₙ₋₁)], R = Δx[f(x₁) + … + f(xₙ)].

Example: f(x) = 4 − x² on [0, 2] with two widths of 1: left sum 4 + 3 = 7, right sum 3 + 0 = 3.

writing a Riemann sum limit as an integral

In lim n→∞ Σ f(a + kΔx)Δx, read off f and a; Δx = (b − a)/n gives b. The limit is ∫ from a to b of f(x) dx.

Example: Σ sin(kπ/n)·(π/n), k = 1 to n, tends to ∫ from 0 to π of sin x dx = 2: Δx = π/n.

Fundamental Theorem of Calculus, derivative form

For continuous f, d/dx [∫ from a to x of f(t) dt] = f(x) on any interval containing a: differentiating an accumulation function returns the integrand at x.

Example: d/dx ∫ from 2 to x of cos(t²) dt = cos(x²), even though cos(t²) has no simple antiderivative.

power rule for integrals

For n ≠ −1, ∫ xⁿ dx = xⁿ⁺¹/(n + 1) + C: raise the exponent by 1, then divide by the new exponent. Rewrite roots and reciprocals as powers first.

Example: ∫ ∛x dx = ∫ x^(1/3) dx = x^(4/3)/(4/3) + C = (3/4)x^(4/3) + C.

changing the limits in a substitution

In a definite integral, convert the limits too: x = a becomes u = u(a) and x = b becomes u = u(b). Then evaluate in u without switching back.

Example: ∫ from 0 to π/2 of sin³x cos x dx: u = sin x with new limits 0 and 1 gives 1⁴/4 = 1/4.

What this area covers

How the exam asks it

Table questions give readings at uneven times; you build a Riemann or trapezoidal sum, give its units and say whether it runs high or low. Graph questions describe f in words (segments, semicircles) and ask about g(x) = ∫ from a to x of f(t) dt. The rest ask you to evaluate an integral or pick an antiderivative; the BC-only ones use parts, linear partial fractions and improper integrals.

Common traps

Trap 1. Missing an asymptote inside the interval. 1/x² is unbounded at x = 0, so ∫ from −1 to 1 of 1/x² dx is improper and diverges. Plugging into −1/x gives −2, impossible for a positive integrand.

Trap 2. Dropping the absolute value in ln|x|. ∫ from −3 to −1 of (1/x) dx = ln|−1| − ln|−3| = −ln 3; without the bars you would take the log of a negative number.

Trap 3. Keeping the x-limits after a substitution. In ∫ from 0 to 1 of 2x(x² + 1)³ dx, u = x² + 1 runs from 1 to 2, giving (16 − 1)/4 = 15/4; using 0 to 1 for u gives 1/4.

Memory trick

LIATE picks u for parts: Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential. Choose u from the type that comes first.

FTC with the chain rule: "plug in the top, times the derivative of the top."

An infinite series adds infinitely many terms, and it has a sum only if its partial sums settle on a limit. This unit tests series for convergence, bounds the error of a partial sum, and builds Taylor polynomials and power series that turn functions into polynomials. It is BC-only and, with Integration and Accumulation, the most heavily weighted unit on the multiple-choice section (15–20% each).

Key terms, with examples

nth partial sum

Sₙ = a₁ + a₂ + … + aₙ adds up the series' first n terms. The partial sums S₁, S₂, S₃, … form a sequence.

Example: For Σ 1/(k(k + 1)), S₃ = 1/2 + 1/6 + 1/12 = 3/4.

nth term test for divergence

If lim n→∞ aₙ ≠ 0, or the limit doesn't exist, then Σ aₙ diverges. If lim aₙ = 0, the test says nothing: the series may still diverge.

Example: Σ n/(3n + 2): the terms approach 1/3, not 0, so the series diverges.

direct comparison test

If 0 ≤ aₙ ≤ bₙ for every n: Σ bₙ converges ⇒ Σ aₙ converges, and Σ aₙ diverges ⇒ Σ bₙ diverges.

Example: 0 < 1/(3ⁿ + 2) < 1/3ⁿ, and Σ 1/3ⁿ is geometric with r = 1/3, so Σ 1/(3ⁿ + 2) converges.

absolute convergence

Σ aₙ converges absolutely if Σ |aₙ| converges. Absolute convergence implies convergence, and any rearranging or regrouping of the terms keeps the same sum.

Example: Σ (−1)ⁿ/n² converges absolutely, because Σ 1/n² is a convergent p-series.

Taylor polynomial

The coefficient of (x − a)ᵏ is f⁽ᵏ⁾(a)/k!: Pₙ(x) = f(a) + f′(a)(x − a) + … + f⁽ⁿ⁾(a)(x − a)ⁿ/n!.

Example: f(2) = 1, f′(2) = 3, f″(2) = −4: P₂(x) = 1 + 3(x − 2) − 2(x − 2)².

power series

Σ from n = 0 to ∞ of aₙ(x − r)ⁿ, centered at x = r. It converges at x = r only, on an interval around r, or for all x.

Example: Σ (x − 3)ⁿ/2ⁿ at x = 4 is Σ (1/2)ⁿ, which equals 2.

Taylor series

The infinite version of the Taylor polynomial: Σ from n = 0 to ∞ of f⁽ⁿ⁾(a)(x − a)ⁿ/n!. Each Taylor polynomial is one of its partial sums.

Example: 1/x centered at 1: f⁽ⁿ⁾(1) = (−1)ⁿn!, so the series is Σ (−1)ⁿ(x − 1)ⁿ.

convergent and divergent series

A series converges to S if its partial sums Sₙ approach S as n → ∞; S is then its sum. If the partial sums have no finite limit, it diverges.

Example: Σ 1/(k(k + 1)): Sₙ = n/(n + 1) approaches 1 as n grows, so the sum is 1.

integral test

If f is positive, continuous and decreasing for x ≥ 1 with f(n) = aₙ, then Σ aₙ and ∫ from 1 to ∞ of f(x) dx both converge or both diverge.

Example: ∫ from 1 to ∞ of 1/(2x + 1) dx = lim b→∞ (1/2)ln((2b + 1)/3) = ∞, so Σ 1/(2n + 1) diverges.

limit comparison test

For positive aₙ and bₙ, if lim aₙ/bₙ = L with 0 < L < ∞, then Σ aₙ and Σ bₙ both converge or both diverge.

Example: aₙ = (2n + 1)/(n³ + 4) behaves like bₙ = 1/n²: the ratio → 2, and Σ 1/n² converges, so Σ aₙ does too.

conditional convergence

Σ aₙ converges conditionally if it converges but Σ |aₙ| diverges. It is usually an alternating series whose all-positive version diverges.

Example: Σ (−1)ⁿ/√n converges by the alternating series test, but Σ 1/√n is a divergent p-series.

Maclaurin polynomial

A Taylor polynomial centered at a = 0, so its terms are f⁽ⁿ⁾(0)xⁿ/n!.

Example: f(x) = √(1 + x): f(0) = 1, f′(0) = 1/2, f″(0) = −1/4, so P₂(x) = 1 + x/2 − x²/8.

What this area covers

How the exam asks it

Expect lists of series with "which converge?" or "which converges conditionally?", a power series and its interval of convergence, a Taylor polynomial built from a table of derivatives, the coefficient of one power in a series, or the most an approximation can be off. Free-response questions may name the test to use and want its conditions checked in words.

Common traps

Trap 1. Forgetting the factorials. The coefficient of (x − a)ⁿ is f⁽ⁿ⁾(a)/n!, so f″(a) = 6 gives 3 as the coefficient of (x − a)², not 6.

Trap 2. Skipping the endpoints. The ratio test says nothing where |x − r| = R, so an interval of convergence isn't finished until each endpoint has been tested.

Trap 3. Using the n = 0 formula on a series that starts at n = 1. Σ from n = 1 to ∞ of 3(1/2)ⁿ has first term 3/2, so its sum is 3, not 6.

Memory trick

Lagrange bound, "one step past": for Pₙ, use the (n + 1)st derivative, the power n + 1 and (n + 1)!.

p-series rhyme: "p bigger than one, the sum gets done."